Answer :

Solution :

The given polynomial is:

ax3 + bx2 + cx + d = 0

Option 1:

At d = 0, the above equation becomes ax3 + bx2 + cx = 0 ⇒ x (ax2 + bx + c) = 0

Now, it is clear that x = 0 is a root for any given set a, b, c.

Therefore, the given statement is correct.

Option 2:

The given polynomial can be expressed as follows.

$x^3+\frac{b}{a}{x}^2+\frac{c}{a}x+\frac{d}{a}=0$

Let the above equation has three equal roots and x = r be a root.

Now,$x^3+\frac{b}{a}{x}^2+\frac{c}{a}x+\frac{d}{a}={\left(x-r\right)}^2=0$

$\Rightarrow x^3+\frac{b}{a}{x}^2+\frac{c}{a}x+\frac{d}{a}=x^3-3r{x}^2+3r^{2}x-r^3=0$

By comparing on both sides,

$\Rightarrow \frac{b}{a}=-3r,\frac{c}{a}=3r^2,\frac{d}{a}=-r^3$

By using the above relations, the conditions to get all the roots equal are b2 = 3ac and bc = 9ad

If we choose the values of a, b, c and d which satisfies the above relations, the roots will be equal, and the corresponding root will be$r=-\frac{b}{3a}$

Therefore, the given statement is incorrect.

Option 3:

The given polynomial is: ax3 + bx2 + cx + d = 0 All the coefficients a, b, c, and d are real

As the coefficients are real, the complex roots must occur in conjugate. The number of possible complex roots are either 0 or 2.

So, no choice of coefficients can make all roots complex.

Therefore, the given statement is incorrect.

Option 4:

c alone cannot ensure that all roots are real.

It depends on all the coefficients.

Therefore, the given statement is incorrect.